Let f: R → R be a twice differentiable function such that (sin x cos y)(f(2 x+2 y)-f(2 x-2 y))=(cos x sin y)(f(2 x+2 y)+f(2 x-2 y)), for all x, y ∈ R. If f′(0)=1/2, then the value of 24 f^(′ ′)((5 π)/3) is :

Solution

Maths sir · 4 steps

Put f(2x+2y) on one side, f(2x-2y) on the other, and the sine addition formulas appear.

1. Regroup the terms. Collect f(2x+2y) on the left and f(2x-2y) on the right. The left coefficient is sin x cos y-cos x sin y=sin(x-y). The right coefficient is sin x cos y+cos x sin y=sin(x+y).

   f(2(x+y))/sin(x+y)=f(2(x-y))/sin(x-y)

2. Find the function. Put u=x+y and v=x-y. These can take any values independently, so f(2u)/(sin u) must be a constant k. Hence f(2u)=k sin u, that is, f(t)=k sin t/2.
3. Use the condition. f'(t)=k/2 cos t/2, so f'(0)=k/2=1/2 gives k=1. Then f(t)=sin t/2 and f''(t)=-1/4 sin t/2.
4. So. f''(5π/3)=-1/4 sin 5π/6=-1/4×1/2=-1/8. So 24f''(5π/3)=-3, which is option (c).

The idea to keep. If a functional equation mixes sin x cos y and cos x sin y, regroup to get sin(x± y). A ratio that is equal for all independent u and v must be a constant.

Answer from the NTA final answer key. Solution by PrepWiser.